in which it is stated as a "folklore" conjecture that the maximal solvable extension of Q is pseudo algebraically closed (this means, in particular, any geometrically connected variety over Q has a point over a solvable extension).

I am curious what evidence there is to support such a conjecture.

In addition, what can be said for the analogous statement for global function fields?

Pete's answer in the linked thread has at least the start of an answer to this question, so potential answerers should read that to avoid duplication. Also: I think "Galois theory of fields" people tend to believe this conjecture more than "rational points on varieties" people, to the extent that those aren't the same people.
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JSENov 22 '09 at 21:42

JSE is correct on both counts. If anyone cares, I didn't say that I believed in the conjecture myself, just that it would be nice if it were true.
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Pete L. ClarkNov 22 '09 at 22:03

2 Answers
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So far as I know, there is no compelling evidence to support this conjecture. (And some leading arithmetic geometers think it is false.) Rather, there are some very interesting consequences of this conjecture, e.g. a solution of the Inverse Galois Problem over Q^{solv}: in other words, for any finite group G, there exists a tower of radical extensions

K_0 = Q, K_1 = K_0(a_0^{1/n_0}) < K_2 = K_1(a_1^{1/n_1}) < ... <= K_n

and a Galois extension L/K_n with Galois group isomorphic to G. It also shows that geometrically irreducible algebraic varieties "acquire rational points" in a very different way from irreducible zero-dimensional varieties (which have a unique minimal
splitting field which need not be solvable).

Of course, interesting things which follow from a conjecture are, if anything, evidence against the truth of the conjecture, although they support the claim that the question is interesting.

There is one impressive result towards this conjecture, namely the Ciperiani-Wiles theorem: let C_{/Q} be a genus one curve with points everywhere locally and semistable Jacobian elliptic curve E. Then C(Q^{solv}) is nonempty.

On the negative side, there is a paper of Ambrus Pal which constructs, for each sufficiently large integer g, a curve C of genus g over a field K which does not admit any points over the maximal solvable extension of K. (Here K is not a number field.)

On the other hand, as far as I know, it is still open to find an absolutely irreducible variety V/Q which fails to have rational points over the maximal metabelian extension of Q, i.e., over (Q^{ab})^{ab}. For some thoughts about this, see

I forgot to address the last part of the question: what about global function fields?

As I alluded to above, there are counterexamples over the function field of a sufficiently complicated ground field, like Q. Of course you mean a finite extension of F_q(T), in which case I think absolutely nothing is known. In particular, I believe the analogue of Ciperiani-Wiles is open here, and might not be a straightforward adaptation, since C-W uses results on the modularity of elliptic curves. This could make a nice thesis problem...but I would talk to Mirela Ciperiani before doing any serious work on it.

I'll respond to this question with another question. Given a curve C/Q, there is a solvable extension K/Q such that C/K has local points everywhere. (At least I think this is is the case; it requires some kind of argument on approximating local solvable extensions by global ones, which I've seen done before.) Some people believe that the Brauer-Manin obstruction is the only obstruction to the Hasse principle for curves. If that were the case, would it say anything about your question?

(At least I think...): Yes. it follows using (i) that every finite extension of a local field is solvable, and (ii) the observation that local extensions of the form X^n - a can be approximated by global extensions of that form (Krasner's Lemma). The assumption about Brauer-Manin would give you that once you've made a solvable extension to get points everywhere locally, then either you have a global point or you have a Brauer obstruction. Are you suggesting that you could then kill the Brauer obstruction by making a further solvable extension? That's interesting...
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Pete L. ClarkNov 22 '09 at 22:21